cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A350309 a(n) = (n+2)*a(n-1) + (n+1)*(A003422(n) - 4)/6 for n > 0 with a(0) = 1.

Original entry on oeis.org

1, 2, 7, 35, 215, 1535, 12455, 113255, 1141415, 12632615, 152341415, 1988514215, 27934434215, 420236744615, 6740662856615, 114841743944615, 2071122598472615, 39418302548552615, 789563088403016615, 16603426141551176615, 365724864314899016615, 8421063413387754056615
Offset: 0

Views

Author

Mikhail Kurkov, Dec 24 2021

Keywords

Crossrefs

Cf. A003422, A006157 (first differences).

Programs

  • Magma
    [n le 2 select n else ((n-1)*(2*n+3)*Self(n-1) - n*(2*n-1)*Self(n-2))/(2*n-3): n in [1..30]]; // G. C. Greubel, Jan 07 2025
    
  • Mathematica
    a[n_]:= a[n]= If[n<2, n+1, (n*(2*n+5)*a[n-1] - (n+1)*(2*n+1)*a[n-2])/(2*n-1)]; (* a = A350309 *)
    Table[a[n], {n,0,30}] (* G. C. Greubel, Jan 07 2025 *)
  • PARI
    lf(n) = sum(k=0, n-1, k!); \\ A003422
    a(n) = if (n, (n+2)*a(n-1) + (n+1)*(lf(n) - 4)/6, 1); \\ Michel Marcus, Jan 11 2022
    
  • Python
    from sage.all import * # remove for SageMath
    @CachedFunction # a = A350309
    def a(n): return n+1 if n<2 else (n*(2*n+5)*a(n-1) - (n+1)*(2*n+1)*a(n-2))//(2*n-1)
    print([a(n) for n in range(31)]) # G. C. Greubel, Jan 07 2025

Formula

Recurrence: (2*n-1)*a(n) = n*(2*n+5)*a(n-1) - (n+1)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Nov 21 2024

A138015 Triangle read by rows, antidiagonals of an array formed by A000012 * A136579. Replace the term "n" in the correlation triangle A003983 with A003422(n).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 4, 10, 4, 2, 1, 1, 2, 4, 10, 10, 4, 2, 1, 1, 2, 4, 10, 34, 10, 4, 2, 1, 1, 2, 4, 10, 34, 34, 10, 4, 2, 1, 1, 2, 4, 10, 34, 154, 34, 10, 4, 2, 1, 1, 2, 4, 10, 34, 154, 154, 34, 10, 4, 2, 1
Offset: 1

Views

Author

Gary W. Adamson, Feb 28 2008

Keywords

Comments

A003422 = (0, 1, 2, 4, 10, 34, 154, ...).
Row sums = A138016: (1, 2, 4, 6, 10, 14, 24, 34, 68, 102, ...).

Examples

			First few rows of the triangle:
  1;
  1, 1;
  1, 2, 1;
  1, 2, 2,  1;
  1, 2, 4,  2,  1;
  1, 2, 4,  4,  2,  1;
  1, 2, 4, 10,  4,  2,  1;
  1, 2, 4, 10, 10,  4,  2, 1;
  1, 2, 4, 10, 34, 10,  4, 2, 1;
  1, 2, 4, 10, 34, 34, 10, 4, 2, 1;
  ...
		

Crossrefs

Extensions

Example corrected by Philippe Deléham, Feb 18 2013

A357145 Decimal expansion of Sum_{n>=1} 1/A003422(n).

Original entry on oeis.org

1, 8, 8, 7, 2, 4, 2, 8, 7, 2, 1, 3, 9, 0, 0, 5, 4, 9, 5, 0, 5, 4, 3, 4, 3, 8, 2, 3, 3, 3, 9, 4, 6, 4, 2, 9, 2, 7, 2, 1, 0, 1, 8, 7, 3, 4, 7, 5, 8, 3, 1, 7, 1, 2, 9, 6, 6, 4, 6, 3, 8, 2, 5, 4, 4, 5, 0, 2, 5, 9, 4, 4, 9, 5, 0, 1, 0, 1, 0, 0, 3, 6, 2, 6, 9, 2, 7, 4, 0, 7, 7, 0, 5, 4, 3, 3, 5, 6, 0, 9, 6, 3, 8, 2, 4, 9
Offset: 1

Views

Author

Hafiz A. Aziz, Sep 15 2022

Keywords

Examples

			1.887242872139005495...
		

Crossrefs

Programs

Extensions

More digits from Alois P. Heinz, Sep 15 2022

A062549 a(n) is the smallest number m such that m! is divisible by the left factorial !n (A003422).

Original entry on oeis.org

1, 2, 4, 5, 17, 11, 23, 2957, 23117, 204557, 2018957, 37313, 296797, 39229, 622561, 1832213, 87616043, 2462521469, 17209100320681, 5207129, 2558906768309, 135954526571, 85412472868721, 3075880875779
Offset: 0

Views

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Nov 22 2001

Keywords

Comments

Conjecture: a(n) is the largest prime factor of !n for all n > 2. - Sean A. Irvine, Apr 01 2023

Examples

			!4 = 1+1+2+6+24 = 34. The smallest value of n such that n! is divisible by 34 is 17!, so a(4)=17
		

A309483 Quotients (!p - Bell(p-1) + 1)/p where p is the n-th prime, !k is Kurepa's left-factorial function (A003422) and Bell(k) is the k-th Bell number (A000110).

Original entry on oeis.org

1, 1, 4, 96, 356540, 39903286, 1312583081304, 356826497344324, 51202108292508282304, 10903333036235662560405182340, 8851961858819132893480466080328, 10341369256681418109100257759613689061054, 20410983764150196478167108200311379711212644128
Offset: 1

Views

Author

Amiram Eldar, Aug 04 2019

Keywords

Comments

Gertsch Hamadene proved that !p == Bell(p-1) - 1 (mod p) for primes p.

Examples

			The 3rd prime is 5, so a(3) = (!5 - Bell(5-1) + 1)/5 = (34 - 15 + 1)/5 = 4.
		

Crossrefs

Programs

  • Mathematica
    quot[p_] := (Sum[k!, {k, 0, p - 1}] - BellB[p - 1] + 1)/p; Table[quot[Prime[i]], {i, 1, 13}]
  • PARI
    a(n) = my(p=prime(n)); (a003422(p) - a000110(p-1) + 1)/p \\ Felix Fröhlich, Aug 04 2019

A376644 Numbers k such that !k * k! + 1 = A003422(k) * A000142(k) + 1 = A061640(k) + 1 is prime.

Original entry on oeis.org

1, 2, 4, 6, 13, 14, 21, 41, 64, 110, 268, 1196, 3083, 5201, 7496, 9698
Offset: 1

Views

Author

Amiram Eldar, Oct 01 2024

Keywords

Comments

a(1)-a(11) were found by Earls (2003-2004).
The corresponding primes are 2, 5, 241, 110881, 3256459844769331201, ... .

Crossrefs

Programs

  • Mathematica
    Select[Range[300], PrimeQ[#! * Sum[k!, {k, 0, #-1}] + 1] &]
  • PARI
    is(k) = isprime(k! * sum(i = 0, k-1, i!) + 1);

Extensions

a(14)-a(16) from Michael S. Branicky, Oct 03 2024

A000142 Factorial numbers: n! = 1*2*3*4*...*n (order of symmetric group S_n, number of permutations of n letters).

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000
Offset: 0

Views

Author

Keywords

Comments

The earliest publication that discusses this sequence appears to be the Sepher Yezirah [Book of Creation], circa AD 300. (See Knuth, also the Zeilberger link.) - N. J. A. Sloane, Apr 07 2014
For n >= 1, a(n) is the number of n X n (0,1) matrices with each row and column containing exactly one entry equal to 1.
This sequence is the BinomialMean transform of A000354. (See A075271 for definition.) - John W. Layman, Sep 12 2002 [This is easily verified from the Paul Barry formula for A000354, by interchanging summations and using the formula: Sum_k (-1)^k C(n-i, k) = KroneckerDelta(i,n). - David Callan, Aug 31 2003]
Number of distinct subsets of T(n-1) elements with 1 element A, 2 elements B, ..., n - 1 elements X (e.g., at n = 5, we consider the distinct subsets of ABBCCCDDDD and there are 5! = 120). - Jon Perry, Jun 12 2003
n! is the smallest number with that prime signature. E.g., 720 = 2^4 * 3^2 * 5. - Amarnath Murthy, Jul 01 2003
a(n) is the permanent of the n X n matrix M with M(i, j) = 1. - Philippe Deléham, Dec 15 2003
Given n objects of distinct sizes (e.g., areas, volumes) such that each object is sufficiently large to simultaneously contain all previous objects, then n! is the total number of essentially different arrangements using all n objects. Arbitrary levels of nesting of objects are permitted within arrangements. (This application of the sequence was inspired by considering leftover moving boxes.) If the restriction exists that each object is able or permitted to contain at most one smaller (but possibly nested) object at a time, the resulting sequence begins 1,2,5,15,52 (Bell Numbers?). Sets of nested wooden boxes or traditional nested Russian dolls come to mind here. - Rick L. Shepherd, Jan 14 2004
From Michael Somos, Mar 04 2004; edited by M. F. Hasler, Jan 02 2015: (Start)
Stirling transform of [2, 2, 6, 24, 120, ...] is A052856 = [2, 2, 4, 14, 76, ...].
Stirling transform of [1, 2, 6, 24, 120, ...] is A000670 = [1, 3, 13, 75, ...].
Stirling transform of [0, 2, 6, 24, 120, ...] is A052875 = [0, 2, 12, 74, ...].
Stirling transform of [1, 1, 2, 6, 24, 120, ...] is A000629 = [1, 2, 6, 26, ...].
Stirling transform of [0, 1, 2, 6, 24, 120, ...] is A002050 = [0, 1, 5, 25, 140, ...].
Stirling transform of (A165326*A089064)(1...) = [1, 0, 1, -1, 8, -26, 194, ...] is [1, 1, 2, 6, 24, 120, ...] (this sequence). (End)
First Eulerian transform of 1, 1, 1, 1, 1, 1... The first Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} e(n, k)s(k), where e(n, k) is a first-order Eulerian number [A008292]. - Ross La Haye, Feb 13 2005
Conjecturally, 1, 6, and 120 are the only numbers which are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
n! is the n-th finite difference of consecutive n-th powers. E.g., for n = 3, [0, 1, 8, 27, 64, ...] -> [1, 7, 19, 37, ...] -> [6, 12, 18, ...] -> [6, 6, ...]. - Bryan Jacobs (bryanjj(AT)gmail.com), Mar 31 2005
a(n+1) = (n+1)! = 1, 2, 6, ... has e.g.f. 1/(1-x)^2. - Paul Barry, Apr 22 2005
Write numbers 1 to n on a circle. Then a(n) = sum of the products of all n - 2 adjacent numbers. E.g., a(5) = 1*2*3 + 2*3*4 + 3*4*5 + 4*5*1 +5*1*2 = 120. - Amarnath Murthy, Jul 10 2005
The number of chains of maximal length in the power set of {1, 2, ..., n} ordered by the subset relation. - Rick L. Shepherd, Feb 05 2006
The number of circular permutations of n letters for n >= 0 is 1, 1, 1, 2, 6, 24, 120, 720, 5040, 40320, ... - Xavier Noria (fxn(AT)hashref.com), Jun 04 2006
a(n) is the number of deco polyominoes of height n (n >= 1; see definitions in the Barcucci et al. references). - Emeric Deutsch, Aug 07 2006
a(n) is the number of partition tableaux of size n. See Steingrimsson/Williams link for the definition. - David Callan, Oct 06 2006
Consider the n! permutations of the integer sequence [n] = 1, 2, ..., n. The i-th permutation consists of ncycle(i) permutation cycles. Then, if the Sum_{i=1..n!} 2^ncycle(i) runs from 1 to n!, we have Sum_{i=1..n!} 2^ncycle(i) = (n+1)!. E.g., for n = 3 we have ncycle(1) = 3, ncycle(2) = 2, ncycle(3) = 1, ncycle(4) = 2, ncycle(5) = 1, ncycle(6) = 2 and 2^3 + 2^2 + 2^1 + 2^2 + 2^1 + 2^2 = 8 + 4 + 2 + 4 + 2 + 4 = 24 = (n+1)!. - Thomas Wieder, Oct 11 2006
a(n) is the number of set partitions of {1, 2, ..., 2n - 1, 2n} into blocks of size 2 (perfect matchings) in which each block consists of one even and one odd integer. For example, a(3) = 6 counts 12-34-56, 12-36-45, 14-23-56, 14-25-36, 16-23-45, 16-25-34. - David Callan, Mar 30 2007
Consider the multiset M = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...] = [1, 2, 2, ..., n x 'n'] and form the set U (where U is a set in the strict sense) of all subsets N (where N may be a multiset again) of M. Then the number of elements |U| of U is equal to (n+1)!. E.g. for M = [1, 2, 2] we get U = [[], [2], [2, 2], [1], [1, 2], [1, 2, 2]] and |U| = 3! = 6. This observation is a more formal version of the comment given already by Rick L. Shepherd, Jan 14 2004. - Thomas Wieder, Nov 27 2007
For n >= 1, a(n) = 1, 2, 6, 24, ... are the positions corresponding to the 1's in decimal expansion of Liouville's constant (A012245). - Paul Muljadi, Apr 15 2008
Triangle A144107 has n! for row sums (given n > 0) with right border n! and left border A003319, the INVERTi transform of (1, 2, 6, 24, ...). - Gary W. Adamson, Sep 11 2008
Equals INVERT transform of A052186 and row sums of triangle A144108. - Gary W. Adamson, Sep 11 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of order-decreasing full transformations (of an n-chain).
a(n-1) is also the number of nilpotent order-decreasing full transformations (of an n-chain). (End)
n! is also the number of optimal broadcast schemes in the complete graph K_{n}, equivalent to the number of binomial trees embedded in K_{n} (see Calin D. Morosan, Information Processing Letters, 100 (2006), 188-193). - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Let S_{n} denote the n-star graph. The S_{n} structure consists of n S_{n-1} structures. This sequence gives the number of edges between the vertices of any two specified S_{n+1} structures in S_{n+2} (n >= 1). - K.V.Iyer, Mar 18 2009
Chromatic invariant of the sun graph S_{n-2}.
It appears that a(n+1) is the inverse binomial transform of A000255. - Timothy Hopper, Aug 20 2009
a(n) is also the determinant of a square matrix, An, whose coefficients are the reciprocals of beta function: a{i, j} = 1/beta(i, j), det(An) = n!. - Enrique Pérez Herrero, Sep 21 2009
The asymptotic expansions of the exponential integrals E(x, m = 1, n = 1) ~ exp(-x)/x*(1 - 1/x + 2/x^2 - 6/x^3 + 24/x^4 + ...) and E(x, m = 1, n = 2) ~ exp(-x)/x*(1 - 2/x + 6/x^2 - 24/x^3 + ...) lead to the factorial numbers. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009
Satisfies A(x)/A(x^2), A(x) = A173280. - Gary W. Adamson, Feb 14 2010
a(n) = G^n where G is the geometric mean of the first n positive integers. - Jaroslav Krizek, May 28 2010
Increasing colored 1-2 trees with choice of two colors for the rightmost branch of nonleaves. - Wenjin Woan, May 23 2011
Number of necklaces with n labeled beads of 1 color. - Robert G. Wilson v, Sep 22 2011
The sequence 1!, (2!)!, ((3!)!)!, (((4!)!)!)!, ..., ((...(n!)!)...)! (n times) grows too rapidly to have its own entry. See Hofstadter.
The e.g.f. of 1/a(n) = 1/n! is BesselI(0, 2*sqrt(x)). See Abramowitz-Stegun, p. 375, 9.3.10. - Wolfdieter Lang, Jan 09 2012
a(n) is the length of the n-th row which is the sum of n-th row in triangle A170942. - Reinhard Zumkeller, Mar 29 2012
Number of permutations of elements 1, 2, ..., n + 1 with a fixed element belonging to a cycle of length r does not depend on r and equals a(n). - Vladimir Shevelev, May 12 2012
a(n) is the number of fixed points in all permutations of 1, ..., n: in all n! permutations, 1 is first exactly (n-1)! times, 2 is second exactly (n-1)! times, etc., giving (n-1)!*n = n!. - Jon Perry, Dec 20 2012
For n >= 1, a(n-1) is the binomial transform of A000757. See Moreno-Rivera. - Luis Manuel Rivera Martínez, Dec 09 2013
Each term is divisible by its digital root (A010888). - Ivan N. Ianakiev, Apr 14 2014
For m >= 3, a(m-2) is the number hp(m) of acyclic Hamiltonian paths in a simple graph with m vertices, which is complete except for one missing edge. For m < 3, hp(m)=0. - Stanislav Sykora, Jun 17 2014
a(n) is the number of increasing forests with n nodes. - Brad R. Jones, Dec 01 2014
The factorial numbers can be calculated by means of the recurrence n! = (floor(n/2)!)^2 * sf(n) where sf(n) are the swinging factorials A056040. This leads to an efficient algorithm if sf(n) is computed via prime factorization. For an exposition of this algorithm see the link below. - Peter Luschny, Nov 05 2016
Treeshelves are ordered (plane) binary (0-1-2) increasing trees where the nodes of outdegree 1 come in 2 colors. There are n! treeshelves of size n, and classical Françon's bijection maps bijectively treeshelves into permutations. - Sergey Kirgizov, Dec 26 2016
Satisfies Benford's law [Diaconis, 1977; Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
a(n) = Sum((d_p)^2), where d_p is the number of standard tableaux in the Ferrers board of the integer partition p and summation is over all integer partitions p of n. Example: a(3) = 6. Indeed, the partitions of 3 are [3], [2,1], and [1,1,1], having 1, 2, and 1 standard tableaux, respectively; we have 1^2 + 2^2 + 1^2 = 6. - Emeric Deutsch, Aug 07 2017
a(n) is the n-th derivative of x^n. - Iain Fox, Nov 19 2017
a(n) is the number of maximum chains in the n-dimensional Boolean cube {0,1}^n in respect to the relation "precedes". It is defined as follows: for arbitrary vectors u, v of {0,1}^n, such that u = (u_1, u_2, ..., u_n) and v = (v_1, v_2, ..., v_n), "u precedes v" if u_i <= v_i, for i=1, 2, ..., n. - Valentin Bakoev, Nov 20 2017
a(n) is the number of shortest paths (for example, obtained by Breadth First Search) between the nodes (0,0,...,0) (i.e., the all-zeros vector) and (1,1,...,1) (i.e., the all-ones vector) in the graph H_n, corresponding to the n-dimensional Boolean cube {0,1}^n. The graph is defined as H_n = (V_n, E_n), where V_n is the set of all vectors of {0,1}^n, and E_n contains edges formed by each pair adjacent vectors. - Valentin Bakoev, Nov 20 2017
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = sigma(gcd(i,j)) for 1 <= i,j <= n. - Bernard Schott, Dec 05 2018
a(n) is also the number of inversion sequences of length n. A length n inversion sequence e_1, e_2, ..., e_n is a sequence of n integers such that 0 <= e_i < i. - Juan S. Auli, Oct 14 2019
The term "factorial" ("factorielle" in French) was coined by the French mathematician Louis François Antoine Arbogast (1759-1803) in 1800. The notation "!" was first used by the French mathematician Christian Kramp (1760-1826) in 1808. - Amiram Eldar, Apr 16 2021
Also the number of signotopes of rank 2, i.e., mappings X:{{1..n} choose 2}->{+,-} such that for any three indices a < b < c, the sequence X(a,b), X(a,c), X(b,c) changes its sign at most once (see Felsner-Weil reference). - Manfred Scheucher, Feb 09 2022
a(n) is also the number of labeled commutative semisimple rings with n elements. As an example the only commutative semisimple rings with 4 elements are F_4 and F_2 X F_2. They both have exactly 2 automorphisms, hence a(4)=24/2+24/2=24. - Paul Laubie, Mar 05 2024
a(n) is the number of extremely unlucky Stirling permutations of order n+1; i.e., the number of Stirling permutations of order n+1 that have exactly one lucky car. - Bridget Tenner, Apr 09 2024

Examples

			There are 3! = 1*2*3 = 6 ways to arrange 3 letters {a, b, c}, namely abc, acb, bac, bca, cab, cba.
Let n = 2. Consider permutations of {1, 2, 3}. Fix element 3. There are a(2) = 2 permutations in each of the following cases: (a) 3 belongs to a cycle of length 1 (permutations (1, 2, 3) and (2, 1, 3)); (b) 3 belongs to a cycle of length 2 (permutations (3, 2, 1) and (1, 3, 2)); (c) 3 belongs to a cycle of length 3 (permutations (2, 3, 1) and (3, 1, 2)). - _Vladimir Shevelev_, May 13 2012
G.f. = 1 + x + 2*x^2 + 6*x^3 + 24*x^4 + 120*x^5 + 720*x^6 + 5040*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 125; also p. 90, ex. 3.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), pars. 448-449.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 64-66.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.1 Symbols Galore, p. 106.
  • Douglas R. Hofstadter, Fluid concepts & creative analogies: computer models of the fundamental mechanisms of thought, Basic Books, 1995, pages 44-46.
  • A. N. Khovanskii. The Application of Continued Fractions and Their Generalizations to Problem in Approximation Theory. Groningen: Noordhoff, Netherlands, 1963. See p. 141 (10.19).
  • D. E. Knuth, The Art of Computer Programming, Vol. 3, Section 5.1.2, p. 23. [From N. J. A. Sloane, Apr 07 2014]
  • J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 693 pp. 90, 297, Ellipses Paris 2004.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • Sepher Yezirah [Book of Creation], circa AD 300. See verse 52.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 2, pages 19-24.
  • D. Stanton and D. White, Constructive Combinatorics, Springer, 1986; see p. 91.
  • Carlo Suares, Sepher Yetsira, Shambhala Publications, 1976. See verse 52.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 102.

Crossrefs

Factorial base representation: A007623.
Complement of A063992. - Reinhard Zumkeller, Oct 11 2008
Cf. A053657, A163176. - Jonathan Sondow, Jul 26 2009
Cf. A173280. - Gary W. Adamson, Feb 14 2010
Boustrophedon transforms: A230960, A230961.
Cf. A233589.
Cf. A245334.
A row of the array in A249026.
Cf. A001013 (multiplicative closure).
For factorials with initial digit d (1 <= d <= 9) see A045509, A045510, A045511, A045516, A045517, A045518, A282021, A045519; A045520, A045521, A045522, A045523, A045524, A045525, A045526, A045527, A045528, A045529.

Programs

  • Axiom
    [factorial(n) for n in 0..10]
    
  • GAP
    List([0..22],Factorial); # Muniru A Asiru, Dec 05 2018
    
  • Haskell
    a000142 :: (Enum a, Num a, Integral t) => t -> a
    a000142 n = product [1 .. fromIntegral n]
    a000142_list = 1 : zipWith (*) [1..] a000142_list
    -- Reinhard Zumkeller, Mar 02 2014, Nov 02 2011, Apr 21 2011
    
  • Julia
    print([factorial(big(n)) for n in 0:28]) # Paul Muljadi, May 01 2024
  • Magma
    a:= func< n | Factorial(n) >; [ a(n) : n in [0..10]];
    
  • Maple
    A000142 := n -> n!; seq(n!,n=0..20);
    spec := [ S, {S=Sequence(Z) }, labeled ]; seq(combstruct[count](spec,size=n), n=0..20);
    # Maple program for computing cycle indices of symmetric groups
    M:=6: f:=array(0..M): f[0]:=1: print(`n= `,0); print(f[0]); f[1]:=x[1]: print(`n= `, 1); print(f[1]); for n from 2 to M do f[n]:=expand((1/n)*add( x[l]*f[n-l],l=1..n)); print(`n= `, n); print(f[n]); od:
    with(combstruct):ZL0:=[S,{S=Set(Cycle(Z,card>0))},labeled]: seq(count(ZL0,size=n),n=0..20); # Zerinvary Lajos, Sep 26 2007
  • Mathematica
    Table[Factorial[n], {n, 0, 20}] (* Stefan Steinerberger, Mar 30 2006 *)
    FoldList[#1 #2 &, 1, Range@ 20] (* Robert G. Wilson v, May 07 2011 *)
    Range[20]! (* Harvey P. Dale, Nov 19 2011 *)
    RecurrenceTable[{a[n] == n*a[n - 1], a[0] == 1}, a, {n, 0, 22}] (* Ray Chandler, Jul 30 2015 *)
  • PARI
    a(n)=prod(i=1, n, i) \\ Felix Fröhlich, Aug 17 2014
    
  • PARI
    {a(n) = if(n<0, 0, n!)}; /* Michael Somos, Mar 04 2004 */
    
  • Python
    for i in range(1, 1000):
        y = i
        for j in range(1, i):
           y *= i - j
        print(y, "\n")
    
  • Python
    import math
    for i in range(1, 1000):
        math.factorial(i)
        print("")
    # Ruskin Harding, Feb 22 2013
    
  • Sage
    [factorial(n) for n in (1..22)] # Giuseppe Coppoletta, Dec 05 2014
    
  • Scala
    (1: BigInt).to(24: BigInt).scanLeft(1: BigInt)( * ) // Alonso del Arte, Mar 02 2019
    

Formula

Exp(x) = Sum_{m >= 0} x^m/m!. - Mohammad K. Azarian, Dec 28 2010
Sum_{i=0..n} (-1)^i * i^n * binomial(n, i) = (-1)^n * n!. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Sum_{i=0..n} (-1)^i * (n-i)^n * binomial(n, i) = n!. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 10 2007
The sequence trivially satisfies the recurrence a(n+1) = Sum_{k=0..n} binomial(n,k) * a(k)*a(n-k). - Robert FERREOL, Dec 05 2009
D-finite with recurrence: a(n) = n*a(n-1), n >= 1. n! ~ sqrt(2*Pi) * n^(n+1/2) / e^n (Stirling's approximation).
a(0) = 1, a(n) = subs(x = 1, (d^n/dx^n)(1/(2-x))), n = 1, 2, ... - Karol A. Penson, Nov 12 2001
E.g.f.: 1/(1-x). - Michael Somos, Mar 04 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*A000522(k)*binomial(n, k) = Sum_{k=0..n} (-1)^(n-k)*(x+k)^n*binomial(n, k). - Philippe Deléham, Jul 08 2004
Binomial transform of A000166. - Ross La Haye, Sep 21 2004
a(n) = Sum_{i=1..n} ((-1)^(i-1) * sum of 1..n taken n - i at a time) - e.g., 4! = (1*2*3 + 1*2*4 + 1*3*4 + 2*3*4) - (1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4) + (1 + 2 + 3 + 4) - 1 = (6 + 8 + 12 + 24) - (2 + 3 + 4 + 6 + 8 + 12) + 10 - 1 = 50 - 35 + 10 - 1 = 24. - Jon Perry, Nov 14 2005
a(n) = (n-1)*(a(n-1) + a(n-2)), n >= 2. - Matthew J. White, Feb 21 2006
1 / a(n) = determinant of matrix whose (i,j) entry is (i+j)!/(i!(j+1)!) for n > 0. This is a matrix with Catalan numbers on the diagonal. - Alexander Adamchuk, Jul 04 2006
Hankel transform of A074664. - Philippe Deléham, Jun 21 2007
For n >= 2, a(n-2) = (-1)^n*Sum_{j=0..n-1} (j+1)*Stirling1(n,j+1). - Milan Janjic, Dec 14 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-x-x^2/(1-3x-4x^2/(1-5x-9x^2/(1-7x-16x^2/(1-9x-25x^2... (continued fraction), hence Hankel transform is A055209.
G.f. of (n+1)! is 1/(1-2x-2x^2/(1-4x-6x^2/(1-6x-12x^2/(1-8x-20x^2... (continued fraction), hence Hankel transform is A059332. (End)
a(n) = Product_{p prime} p^(Sum_{k > 0} floor(n/p^k)) by Legendre's formula for the highest power of a prime dividing n!. - Jonathan Sondow, Jul 24 2009
a(n) = A053657(n)/A163176(n) for n > 0. - Jonathan Sondow, Jul 26 2009
It appears that a(n) = (1/0!) + (1/1!)*n + (3/2!)*n*(n-1) + (11/3!)*n*(n-1)*(n-2) + ... + (b(n)/n!)*n*(n-1)*...*2*1, where a(n) = (n+1)! and b(n) = A000255. - Timothy Hopper, Aug 12 2009
Sum_{n >= 0} 1/a(n) = e. - Jaume Oliver Lafont, Mar 03 2009
a(n) = a(n-1)^2/a(n-2) + a(n-1), n >= 2. - Jaume Oliver Lafont, Sep 21 2009
a(n) = Gamma(n+1). - Enrique Pérez Herrero, Sep 21 2009
a(n) = A173333(n,1). - Reinhard Zumkeller, Feb 19 2010
a(n) = A_{n}(1) where A_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n) = n*(2*a(n-1) - (n-1)*a(n-2)), n > 1. - Gary Detlefs, Sep 16 2010
1/a(n) = -Sum_{k=1..n+1} (-2)^k*(n+k+2)*a(k)/(a(2*k+1)*a(n+1-k)). - Groux Roland, Dec 08 2010
From Vladimir Shevelev, Feb 21 2011: (Start)
a(n) = Product_{p prime, p <= n} p^(Sum_{i >= 1} floor(n/p^i)).
The infinitary analog of this formula is: a(n) = Product_{q terms of A050376 <= n} q^((n)_q), where (n)_q denotes the number of those numbers <= n for which q is an infinitary divisor (for the definition see comment in A037445). (End)
The terms are the denominators of the expansion of sinh(x) + cosh(x). - Arkadiusz Wesolowski, Feb 03 2012
G.f.: 1 / (1 - x / (1 - x / (1 - 2*x / (1 - 2*x / (1 - 3*x / (1 - 3*x / ... )))))). - Michael Somos, May 12 2012
G.f. 1 + x/(G(0)-x) where G(k) = 1 - (k+1)*x/(1 - x*(k+2)/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 14 2012
G.f.: W(1,1;-x)/(W(1,1;-x) - x*W(1,2;-x)), where W(a,b,x) = 1 - a*b*x/1! + a*(a+1)*b*(b+1)*x^2/2! - ... + a*(a+1)*...*(a+n-1)*b*(b+1)*...*(b+n-1)*x^n/n! + ...; see [A. N. Khovanskii, p. 141 (10.19)]. - Sergei N. Gladkovskii, Aug 15 2012
From Sergei N. Gladkovskii, Dec 26 2012: (Start)
G.f.: A(x) = 1 + x/(G(0) - x) where G(k) = 1 + (k+1)*x - x*(k+2)/G(k+1); (continued fraction).
Let B(x) be the g.f. for A051296, then A(x) = 2 - 1/B(x). (End)
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (2*k+1)/(1-x/(x - 1/(1 - (2*k+2)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*(1 - G(0))/(sqrt(x)-x) where G(k) = 1 - (k+1)*sqrt(x)/(1-sqrt(x)/(sqrt(x)-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 25 2013
G.f.: 1 + x/G(0) where G(k) = 1 - x*(k+2)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013
a(n) = det(S(i+1, j), 1 <= i, j <=n ), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 04 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
G.f.: G(0), where G(k) = 1 + x*(2*k+1)/(1 - x*(2*k+2)/(x*(2*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 07 2013
a(n) = P(n-1, floor(n/2)) * floor(n/2)! * (n - (n-2)*((n+1) mod 2)), where P(n, k) are the k-permutations of n objects, n > 0. - Wesley Ivan Hurt, Jun 07 2013
a(n) = a(n-2)*(n-1)^2 + a(n-1), n > 1. - Ivan N. Ianakiev, Jun 18 2013
a(n) = a(n-2)*(n^2-1) - a(n-1), n > 1. - Ivan N. Ianakiev, Jun 30 2013
G.f.: 1 + x/Q(0), m=+2, where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
a(n) = A245334(n,n). - Reinhard Zumkeller, Aug 31 2014
a(n) = Product_{i = 1..n} A014963^floor(n/i) = Product_{i = 1..n} A003418(floor(n/i)). - Matthew Vandermast, Dec 22 2014
a(n) = round(Sum_{k>=1} log(k)^n/k^2), for n>=1, which is related to the n-th derivative of the Riemann zeta function at x=2 as follows: round((-1)^n * zeta^(n)(2)). Also see A073002. - Richard R. Forberg, Dec 30 2014
a(n) ~ Sum_{j>=0} j^n/e^j, where e = A001113. When substituting a generic variable for "e" this infinite sum is related to Eulerian polynomials. See A008292. This approximation of n! is within 0.4% at n = 2. See A255169. Accuracy, as a percentage, improves rapidly for larger n. - Richard R. Forberg, Mar 07 2015
a(n) = Product_{k=1..n} (C(n+1, 2)-C(k, 2))/(2*k-1); see Masanori Ando link. - Michel Marcus, Apr 17 2015
Sum_{n>=0} a(n)/(a(n + 1)*a(n + 2)) = Sum_{n>=0} 1/((n + 2)*(n + 1)^2*a(n)) = 2 - exp(1) - gamma + Ei(1) = 0.5996203229953..., where gamma = A001620, Ei(1) = A091725. - Ilya Gutkovskiy, Nov 01 2016
a(2^n) = 2^(2^n - 1) * 1!! * 3!! * 7!! * ... * (2^n - 1)!!. For example, 16! = 2^15*(1*3)*(1*3*5*7)*(1*3*5*7*9*11*13*15) = 20922789888000. - Peter Bala, Nov 01 2016
a(n) = sum(prod(B)), where the sum is over all subsets B of {1,2,...,n-1} and where prod(B) denotes the product of all the elements of set B. If B is a singleton set with element b, then we define prod(B)=b, and, if B is the empty set, we define prod(B) to be 1. For example, a(4)=(1*2*3)+(1*2)+(1*3)+(2*3)+(1)+(2)+(3)+1=24. - Dennis P. Walsh, Oct 23 2017
Sum_{n >= 0} 1/(a(n)*(n+2)) = 1. - Multiplying the denominator by (n+2) in Jaume Oliver Lafont's entry above creates a telescoping sum. - Fred Daniel Kline, Nov 08 2020
O.g.f.: Sum_{k >= 0} k!*x^k = Sum_{k >= 0} (k+y)^k*x^k/(1 + (k+y)*x)^(k+1) for arbitrary y. - Peter Bala, Mar 21 2022
E.g.f.: 1/(1 + LambertW(-x*exp(-x))) = 1/(1-x), see A258773. -(1/x)*substitute(z = x*exp(-x), z*(d/dz)LambertW(-z)) = 1/(1 - x). See A075513. Proof: Use the compositional inverse (x*exp(-x))^[-1] = -LambertW(-z). See A000169 or A152917, and Richard P. Stanley: Enumerative Combinatorics, vol. 2, p. 37, eq. (5.52). - Wolfdieter Lang, Oct 17 2022
Sum_{k >= 1} 1/10^a(k) = A012245 (Liouville constant). - Bernard Schott, Dec 18 2022
From David Ulgenes, Sep 19 2023: (Start)
1/a(n) = (e/(2*Pi*n)*Integral_{x=-oo..oo} cos(x-n*arctan(x))/(1+x^2)^(n/2) dx). Proof: take the real component of Laplace's integral for 1/Gamma(x).
a(n) = Integral_{x=0..1} e^(-t)*LerchPhi(1/e, -n, t) dt. Proof: use the relationship Gamma(x+1) = Sum_{n >= 0} Integral_{t=n..n+1} e^(-t)t^x dt = Sum_{n >= 0} Integral_{t=0..1} e^(-(t+n))(t+n)^x dt and interchange the order of summation and integration.
Conjecture: a(n) = 1/(2*Pi)*Integral_{x=-oo..oo}(n+i*x+1)!/(i*x+1)-(n+i*x-1)!/(i*x-1)dx. (End)
a(n) = floor(b(n)^n / (floor(((2^b(n) + 1) / 2^n)^b(n)) mod 2^b(n))), where b(n) = (n + 1)^(n + 2) = A007778(n+1). Joint work with Mihai Prunescu. - Lorenzo Sauras Altuzarra, Oct 18 2023
a(n) = e^(Integral_{x=1..n+1} Psi(x) dx) where Psi(x) is the digamma function. - Andrea Pinos, Jan 10 2024
a(n) = Integral_{x=0..oo} e^(-x^(1/n)) dx, for n > 0. - Ridouane Oudra, Apr 20 2024
O.g.f.: N(x) = hypergeometric([1,1], [], x) = LaplaceTransform(x/(1-x))/x, satisfying x^2*N'(x) + (x-1)*N(x) + 1 = 0, with N(0) = 1. - Wolfdieter Lang, May 31 2025

A000110 Bell or exponential numbers: number of ways to partition a set of n labeled elements.

Original entry on oeis.org

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346, 445958869294805289, 4638590332229999353, 49631246523618756274
Offset: 0

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Author

Keywords

Comments

The leading diagonal of its difference table is the sequence shifted, see Bernstein and Sloane (1995). - N. J. A. Sloane, Jul 04 2015
Also the number of equivalence relations that can be defined on a set of n elements. - Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005
a(n) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. Adjacent regions cannot have the same color. - David W. Wilson, Feb 22 2005
If an integer is squarefree and has n distinct prime factors then a(n) is the number of ways of writing it as a product of its divisors. - Amarnath Murthy, Apr 23 2001
Consider rooted trees of height at most 2. Letting each tree 'grow' into the next generation of n means we produce a new tree for every node which is either the root or at height 1, which gives the Bell numbers. - Jon Perry, Jul 23 2003
Begin with [1,1] and follow the rule that [1,k] -> [1,k+1] and [1,k] k times, e.g., [1,3] is transformed to [1,4], [1,3], [1,3], [1,3]. Then a(n) is the sum of all components: [1,1] = 2; [1,2], [1,1] = 5; [1,3], [1,2], [1,2], [1,2], [1,1] = 15; etc. - Jon Perry, Mar 05 2004
Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (e.g., 'abba') such that the leftmost letter is always 'a' and no letter may be greater than one more than the greatest letter to its left. Thus 'aac' is not valid since 'c' is more than one greater than 'a'. For example, a(3)=5 because there are 5 rhyme schemes: aaa, aab, aba, abb, abc; also see example by Neven Juric. - Bill Blewett, Mar 23 2004
In other words, number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= 1 + max(prefix) for k >= 1, see example (cf. A080337 and A189845). - Joerg Arndt, Apr 30 2011
Number of partitions of {1, ..., n+1} into subsets of nonconsecutive integers, including the partition 1|2|...|n+1. E.g., a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers, namely, 13|24, 13|2|4, 14|2|3, 1|24|3, 1|2|3|4. - Augustine O. Munagi, Mar 20 2005
Triangle (addition) scheme to produce terms, derived from the recurrence, from Oscar Arevalo (loarevalo(AT)sbcglobal.net), May 11 2005:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
... [This is Aitken's array A011971]
With P(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..P(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (1/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
a(n+1) is the number of binary relations on an n-element set that are both symmetric and transitive. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
If the rule from Jon Perry, Mar 05 2004, is used, then a(n-1) = [number of components used to form a(n)] / 2. - Daniel Kuan (dkcm(AT)yahoo.com), Feb 19 2006
a(n) is the number of functions f from {1,...,n} to {1,...,n,n+1} that satisfy the following two conditions for all x in the domain: (1) f(x) > x; (2) f(x)=n+1 or f(f(x))=n+1. E.g., a(3)=5 because there are exactly five functions that satisfy the two conditions: f1={(1,4),(2,4),(3,4)}, f2={(1,4),(2,3),(3,4)}, f3={(1,3),(2,4),(3,4)}, f4={(1,2),(2,4),(3,4)} and f5={(1,3),(2,3),(3,4)}. - Dennis P. Walsh, Feb 20 2006
Number of asynchronic siteswap patterns of length n which have no zero-throws (i.e., contain no 0's) and whose number of orbits (in the sense given by Allen Knutson) is equal to the number of balls. E.g., for n=4, the condition is satisfied by the following 15 siteswaps: 4444, 4413, 4242, 4134, 4112, 3441, 2424, 1344, 2411, 1313, 1241, 2222, 3131, 1124, 1111. Also number of ways to choose n permutations from identity and cyclic permutations (1 2), (1 2 3), ..., (1 2 3 ... n) so that their composition is identity. For n=3 we get the following five: id o id o id, id o (1 2) o (1 2), (1 2) o id o (1 2), (1 2) o (1 2) o id, (1 2 3) o (1 2 3) o (1 2 3). (To see the bijection, look at Ehrenborg and Readdy paper.) - Antti Karttunen, May 01 2006
a(n) is the number of permutations on [n] in which a 3-2-1 (scattered) pattern occurs only as part of a 3-2-4-1 pattern. Example: a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" reference a(n) = number of permutation tableaux of size n (A000142) whose first row contains no 0's. Example: a(3)=5 counts {{}, {}, {}}, {{1}, {}}, {{1}, {0}}, {{1}, {1}}, {{1, 1}}. - David Callan, Oct 07 2006
From Gottfried Helms, Mar 30 2007: (Start)
This sequence is also the first column in the matrix-exponential of the (lower triangular) Pascal-matrix, scaled by exp(-1): PE = exp(P) / exp(1) =
1
1 1
2 2 1
5 6 3 1
15 20 12 4 1
52 75 50 20 5 1
203 312 225 100 30 6 1
877 1421 1092 525 175 42 7 1
First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamard-product Toeplitz(A000110) (X) P:
1
1 1
2 1 1
5 2 1 1
15 5 2 1 1 (X) P
52 15 5 2 1 1
203 52 15 5 2 1 1
877 203 52 15 5 2 1 1
(End)
The terms can also be computed with finite steps and precise integer arithmetic. Instead of exp(P)/exp(1) one can compute A = exp(P - I) where I is the identity-matrix of appropriate dimension since (P-I) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the row-index starting at 1. - Gottfried Helms, Apr 10 2007
When n is prime, a(n) == 2 (mod n), but the converse is not always true. Define a Bell pseudoprime to be a composite number n such that a(n) == 2 (mod n). W. F. Lunnon recently found the Bell pseudoprimes 21361 = 41*521 and C46 = 3*23*16218646893090134590535390526854205539989357 and conjectured that Bell pseudoprimes are extremely scarce. So the second Bell pseudoprime is unlikely to be known with certainty in the near future. I confirmed that 21361 is the first. - David W. Wilson, Aug 04 2007 and Sep 24 2007
This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314. - Tom Copeland, Oct 21 2007
Starting (1, 2, 5, 15, 52, ...), equals row sums and right border of triangle A136789. Also row sums of triangle A136790. - Gary W. Adamson, Jan 21 2008
This is the exponential transform of A000012. - Thomas Wieder, Sep 09 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of idempotent order-decreasing full transformations (of an n-chain).
a(n) is also the number of nilpotent partial one-one order-decreasing transformations (of an n-chain).
a(n+1) is also the number of partial one-one order-decreasing transformations (of an n-chain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of n-pattern sequences [Cooper & Kennedy]. An n-pattern sequence is a sequence of integers (a_1,...,a_n) such that a_i = i or a_i = a_j for some j < i. For example, Bell(3) = 5 since the 3-pattern sequences are (1,1,1), (1,1,3), (1,2,1), (1,2,2) and (1,2,3).
Bell(n) is the number of sequences of positive integers (N_1,...,N_n) of length n such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (see the comment by B. Blewett above). It is interesting to note that if we strengthen the latter condition to N_(i+1) <= 1 + N_i we get the Catalan numbers A000108 instead of the Bell numbers.
(End)
Equals the eigensequence of Pascal's triangle, A007318; and starting with offset 1, = row sums of triangles A074664 and A152431. - Gary W. Adamson, Dec 04 2008
The entries f(i, j) in the exponential of the infinite lower-triangular matrix of binomial coefficients b(i, j) are f(i, j) = b(i, j) e a(i - j). - David Pasino, Dec 04 2008
Equals lim_{k->oo} A071919^k. - Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1-x-exp(-x)), the eigensequence of A007318^(-1). Starting with offset 1 = A154108 convolved with (1,2,3,...) = row sums of triangle A154109. - Gary W. Adamson, Jan 04 2009
Repeated iterates of (binomial transform of [1,0,0,0,...]) will converge upon (1, 2, 5, 15, 52, ...) when each result is prefaced with a "1"; such that the final result is the fixed limit: (binomial transform of [1,1,2,5,15,...]) = (1,2,5,15,52,...). - Gary W. Adamson, Jan 14 2009
From Karol A. Penson, May 03 2009: (Start)
Relation between the Bell numbers B(n) and the n-th derivative of 1/Gamma(1+x) evaluated at x=1:
a) produce a number of such derivatives through seq(subs(x=1, simplify((d^n/dx^n)GAMMA(1+x)^(-1))), n=1..5);
b) leave them expressed in terms of digamma (Psi(k)) and polygamma (Psi(k,n)) functions and unevaluated;
Examples of such expressions, for n=1..5, are:
n=1: -Psi(1),
n=2: -(-Psi(1)^2 + Psi(1,1)),
n=3: -Psi(1)^3 + 3*Psi(1)*Psi(1,1) - Psi(2,1),
n=4: -(-Psi(1)^4 + 6*Psi(1)^2*Psi(1,1) - 3*Psi(1,1)^2 - 4*Psi(1)*Psi(2,1) + Psi(3, 1)),
n=5: -Psi(1)^5 + 10*Psi(1)^3*Psi(1,1) - 15*Psi(1)*Psi(1,1)^2 - 10*Psi(1)^2*Psi(2,1) + 10*Psi(1,1)*Psi(2,1) + 5*Psi(1)*Psi(3,1) - Psi(4,1);
c) for a given n, read off the sum of absolute values of coefficients of every term involving digamma or polygamma functions.
This sum is equal to B(n). Examples: B(1)=1, B(2)=1+1=2, B(3)=1+3+1=5, B(4)=1+6+3+4+1=15, B(5)=1+10+15+10+10+5+1=52;
d) Observe that this decomposition of the Bell number B(n) apparently does not involve the Stirling numbers of the second kind explicitly.
(End)
The numbers given above by Penson lead to the multinomial coefficients A036040. - Johannes W. Meijer, Aug 14 2009
Column 1 of A162663. - Franklin T. Adams-Watters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n-1)!)/(n-1)! = a(0) + a(1)/n + a(2)/n^2 + ... and (0!+1!+2!+...+n!)/n! = 1 + a(0)/n + a(1)/n^2 + a(2)/n^3 + .... - Michael Somos, Jun 28 2009
Starting with offset 1 = row sums of triangle A165194. - Gary W. Adamson, Sep 06 2009
a(n+1) = A165196(2^n); where A165196 begins: (1, 2, 4, 5, 7, 12, 14, 15, ...). such that A165196(2^3) = 15 = A000110(4). - Gary W. Adamson, Sep 06 2009
The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m >= -1, which for m=-1 dates back to Euler, is related to the Bell numbers. We discovered that g(x=1,m) = (-1)^m * (A040027(m) - A000110(m+1) * A073003). We observe that A073003 is Gompertz's constant and that A040027 was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009
a(n) = E(X^n), i.e., the n-th moment about the origin of a random variable X that has a Poisson distribution with (rate) parameter, lambda = 1. - Geoffrey Critzer, Nov 30 2009
Let A000110 = S(x), then S(x) = A(x)/A(x^2) when A(x) = A173110; or (1, 1, 2, 5, 15, 52, ...) = (1, 1, 3, 6, 20, 60, ...) / (1, 0, 1, 0, 3, 0, 6, 0, 20, ...). - Gary W. Adamson, Feb 09 2010
The Bell numbers serve as the upper limit for the number of distinct homomorphic images from any given finite universal algebra. Every algebra homomorphism is determined by its kernel, which must be a congruence relation. As the number of possible congruence relations with respect to a finite universal algebra must be a subset of its possible equivalence classes (given by the Bell numbers), it follows naturally. - Max Sills, Jun 01 2010
For a proof of the o.g.f. given in the R. Stephan comment see, e.g., the W. Lang link under A071919. - Wolfdieter Lang, Jun 23 2010
Let B(x) = (1 + x + 2x^2 + 5x^3 + ...). Then B(x) is satisfied by A(x)/A(x^2) where A(x) = polcoeff A173110: (1 + x + 3x^2 + 6x^3 + 20x^4 + 60x^5 + ...) = B(x) * B(x^2) * B(x^4) * B(x^8) * .... - Gary W. Adamson, Jul 08 2010
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color without choosing any two colors the same positive number of times. (See related comments for A000108, A008277, A016098.) - Matthew Vandermast, Nov 22 2010
A binary counter with faulty bits starts at value 0 and attempts to increment by 1 at each step. Each bit that should toggle may or may not do so. a(n) is the number of ways that the counter can have the value 0 after n steps. E.g., for n=3, the 5 trajectories are 0,0,0,0; 0,1,0,0; 0,1,1,0; 0,0,1,0; 0,1,3,0. - David Scambler, Jan 24 2011
No Bell number is divisible by 8, and no Bell number is congruent to 6 modulo 8; see Theorem 6.4 and Table 1.7 in Lunnon, Pleasants and Stephens. - Jon Perry, Feb 07 2011, clarified by Eric Rowland, Mar 26 2014
a(n+1) is the number of (symmetric) positive semidefinite n X n 0-1 matrices. These correspond to equivalence relations on {1,...,n+1}, where matrix element M[i,j] = 1 if and only if i and j are equivalent to each other but not to n+1. - Robert Israel, Mar 16 2011
a(n) is the number of monotonic-labeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotonic-labeled if the label of any parent vertex is greater than the label of any offspring vertex. See link "Counting forests with Stirling and Bell numbers". - Dennis P. Walsh, Nov 11 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000772 and A094198. - Peter Bala, Nov 25 2011
B(n) counts the length n+1 rhyme schemes without repetitions. E.g., for n=2 there are 5 rhyme schemes of length 3 (aaa, aab, aba, abb, abc), and the 2 without repetitions are aba, abc. This is basically O. Munagi's result that the Bell numbers count partitions into subsets of nonconsecutive integers (see comment above dated Mar 20 2005). - Eric Bach, Jan 13 2012
Right and left borders and row sums of A212431 = A000110 or a shifted variant. - Gary W. Adamson, Jun 21 2012
Number of maps f: [n] -> [n] where f(x) <= x and f(f(x)) = f(x) (projections). - Joerg Arndt, Jan 04 2013
Permutations of [n] avoiding any given one of the 8 dashed patterns in the equivalence classes (i) 1-23, 3-21, 12-3, 32-1, and (ii) 1-32, 3-12, 21-3, 23-1. (See Claesson 2001 reference.) - David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e-1) = 5.57494152476..., see A234473. - Richard R. Forberg, Dec 26 2013 (This is the e.g.f. for x=1. - Wolfdieter Lang, Feb 02 2015)
Sum_{j=0..n} binomial(n,j)*a(j) = (1/e)*Sum_{k>=0} (k+1)^n/k! = (1/e) Sum_{k=1..oo} k^(n+1)/k! = a(n+1), n >= 0, using the Dobinski formula. See the comment by Gary W. Adamson, Dec 04 2008 on the Pascal eigensequence. - Wolfdieter Lang, Feb 02 2015
In fact it is not really an eigensequence of the Pascal matrix; rather the Pascal matrix acts on the sequence as a shift. It is an eigensequence (the unique eigensequence with eigenvalue 1) of the matrix derived from the Pascal matrix by adding at the top the row [1, 0, 0, 0 ...]. The binomial sum formula may be derived from the definition in terms of partitions: label any element X of a set S of N elements, and let X(k) be the number of subsets of S containing X with k elements. Since each subset has a unique coset, the number of partitions p(N) of S is given by p(N) = Sum_{k=1..N} (X(k) p(N-k)); trivially X(k) = N-1 choose k-1. - Mason Bogue, Mar 20 2015
a(n) is the number of ways to nest n matryoshkas (Russian nesting dolls): we may identify {1, 2, ..., n} with dolls of ascending sizes and the sets of a set partition with stacks of dolls. - Carlo Sanna, Oct 17 2015
Number of permutations of [n] where the initial elements of consecutive runs of increasing elements are in decreasing order. a(4) = 15: `1234, `2`134, `23`14, `234`1, `24`13, `3`124, `3`2`14, `3`24`1, `34`12, `34`2`1, `4`123, `4`2`13, `4`23`1, `4`3`12, `4`3`2`1. - Alois P. Heinz, Apr 27 2016
Taking with alternating signs, the Bell numbers are the coefficients in the asymptotic expansion (Ramanujan): (-1)^n*(A000166(n) - n!/exp(1)) ~ 1/n - 2/n^2 + 5/n^3 - 15/n^4 + 52/n^5 - 203/n^6 + O(1/n^7). - Vladimir Reshetnikov, Nov 10 2016
Number of treeshelves avoiding pattern T231. See A278677 for definitions and examples. - Sergey Kirgizov, Dec 24 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. - N. J. A. Sloane, Feb 09 2017
a(n) = Sum(# of standard immaculate tableaux of shape m, m is a composition of n), where this sum is over all integer compositions m of n > 0. This formula is easily seen to hold by identifying standard immaculate tableaux of size n with set partitions of { 1, 2, ..., n }. For example, if we sum over integer compositions of 4 lexicographically, we see that 1+1+2+1+3+3+3+1 = 15 = A000110(4). - John M. Campbell, Jul 17 2017
a(n) is also the number of independent vertex sets (and vertex covers) in the (n-1)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017
Even-numbered entries represent the numbers of configurations of identity and non-identity for alleles of a gene in n diploid individuals with distinguishable maternal and paternal alleles. - Noah A Rosenberg, Jan 28 2019
Number of partial equivalence relations (PERs) on a set with n elements (offset=1), i.e., number of symmetric, transitive (not necessarily reflexive) relations. The idea is to add a dummy element D to the set, and then take equivalence relations on the result; anything equivalent to D is then removed for the partial equivalence relation. - David Spivak, Feb 06 2019
Number of words of length n+1 with no repeated letters, when letters are unlabeled. - Thomas Anton, Mar 14 2019
Named by Becker and Riordan (1948) after the Scottish-American mathematician and writer Eric Temple Bell (1883 - 1960). - Amiram Eldar, Dec 04 2020
Also the number of partitions of {1,2,...,n+1} with at most one n+1 singleton. E.g., a(3)=5: {13|24, 12|34, 123|4, 14|23, 1234}. - Yuchun Ji, Dec 21 2020
a(n) is the number of sigma algebras on the set of n elements. Note that each sigma algebra is generated by a partition of the set. For example, the sigma algebra generated by the partition {{1}, {2}, {3,4}} is {{}, {1}, {2}, {1,2}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. - Jianing Song, Apr 01 2021
a(n) is the number of P_3-free graphs on n labeled nodes. - M. Eren Kesim, Jun 04 2021
a(n) is the number of functions X:([n] choose 2) -> {+,-} such that for any ordered 3-tuple abc we have X(ab)X(ac)X(bc) not in {+-+,++-,-++}. - Robert Lauff, Dec 09 2022
From Manfred Boergens, Mar 11 2025: (Start)
The partitions in the definition can be described as disjoint covers of the set. "Covers" in general give rise to the following amendments:
For disjoint covers which may include one empty set see A186021.
For arbitrary (including non-disjoint) covers see A003465.
For arbitrary (including non-disjoint) covers which may include one empty set see A000371. (End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 + 4140*x^8 + ...
From Neven Juric, Oct 19 2009: (Start)
The a(4)=15 rhyme schemes for n=4 are
  aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, abcd
The a(5)=52 rhyme schemes for n=5 are
  aaaaa, aaaab, aaaba, aaabb, aaabc, aabaa, aabab, aabac, aabba, aabbb, aabbc, aabca, aabcb, aabcc, aabcd, abaaa, abaab, abaac, ababa, ababb, ababc, abaca, abacb, abacc, abacd, abbaa, abbab, abbac, abbba, abbbb, abbbc, abbca, abbcb, abbcc, abbcd, abcaa, abcab, abcac, abcad, abcba, abcbb, abcbc, abcbd, abcca, abccb, abccc, abccd, abcda, abcdb, abcdc, abcdd, abcde
(End)
From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings (RGS):
For n=0 there is one empty string;
for n=1 there is one string [0];
for n=2 there are 2 strings [00], [01];
for n=3 there are a(3)=5 strings [000], [001], [010], [011], and [012];
for n=4 there are a(4)=15 strings
1: [0000], 2: [0001], 3: [0010], 4: [0011], 5: [0012], 6: [0100], 7: [0101], 8: [0102], 9: [0110], 10: [0111], 11: [0112], 12: [0120], 13: [0121], 14: [0122], 15: [0123].
These are one-to-one with the rhyme schemes (identify a=0, b=1, c=2, etc.).
(End)
Consider the set S = {1, 2, 3, 4}. The a(4) = 1 + 3 + 6 + 4 + 1 = 15 partitions are: P1 = {{1}, {2}, {3}, {4}}; P21 .. P23 = {{a,4}, S\{a,4}} with a = 1, 2, 3; P24 .. P29 = {{a}, {b}, S\{a,b}} with 1 <= a < b <= 4;  P31 .. P34 = {S\{a}, {a}} with a = 1 .. 4; P4 = {S}. See the Bottomley link for a graphical illustration. - _M. F. Hasler_, Oct 26 2017
		

References

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Crossrefs

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Equals row sums of triangle A152432.
Row sums, right and left borders of A212431.
A diagonal of A011971. - N. J. A. Sloane, Jul 31 2012
Diagonal of A102661. - Manfred Boergens, Mar 11 2025
Cf. A054767 (period of this sequence mod n).
Row sums are A048993. - Wolfdieter Lang, Oct 16 2014
Sequences in the Erné (1974) paper: A000110, A000798, A001035, A001927, A001929, A006056, A006057, A006058, A006059.
Bell polynomials B(n,x): A001861 (x=2), A027710 (x=3), A078944 (x=4), A144180 (x=5), A144223 (x=6), A144263 (x=7), A221159 (x=8).
Cf. A243991 (sum of reciprocals), A085686 (inv. Euler Transf.).

Programs

  • Haskell
    type N = Integer
    n_partitioned_k :: N -> N -> N
    1 `n_partitioned_k` 1 = 1
    1 `n_partitioned_k` _ = 0
    n `n_partitioned_k` k = k * (pred n `n_partitioned_k` k) + (pred n `n_partitioned_k` pred k)
    n_partitioned :: N -> N
    n_partitioned 0 = 1
    n_partitioned n = sum $ map (\k -> n `n_partitioned_k` k) $ [1 .. n]
    -- Felix Denis, Oct 16 2012
    
  • Haskell
    a000110 = sum . a048993_row -- Reinhard Zumkeller, Jun 30 2013
    
  • Julia
    function a(n)
        t = [zeros(BigInt, n+1) for _ in 1:n+1]
        t[1][1] = 1
        for i in 2:n+1
            t[i][1] = t[i-1][i-1]
            for j in 2:i
                t[i][j] = t[i-1][j-1] + t[i][j-1]
            end
        end
        return [t[i][1] for i in 1:n+1]
    end
    print(a(28)) # Paul Muljadi, May 07 2024
    
  • Magma
    [Bell(n): n in [0..40]]; // Vincenzo Librandi, Feb 07 2011
    
  • Maple
    A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1,i)*A000110(n-1-i),i=0..n-1); fi; end: # version 1
    A := series(exp(exp(x)-1),x,60): A000110 := n->n!*coeff(A,x,n): # version 2
    A000110:= n-> add(Stirling2(n, k), k=0..n): seq(A000110(n), n=0..22); # version 3, from Zerinvary Lajos, Jun 28 2007
    A000110 := n -> combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
    spec:= [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]: G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22);  # version 5, Zerinvary Lajos, Dec 16 2007
    BellList := proc(m) local A, P, n; A := [1, 1]; P := [1]; for n from 1 to m - 2 do
    P := ListTools:-PartialSums([A[-1], op(P)]); A := [op(A), P[-1]] od; A end: BellList(29); # Peter Luschny, Mar 24 2022
  • Mathematica
    f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v *)
    Table[BellB[n], {n, 0, 40}] (* Harvey P. Dale, Mar 01 2011 *)
    B[0] = 1; B[n_] := 1/E Sum[k^(n - 1)/(k-1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
    BellB[Range[0,40]] (* Eric W. Weisstein, Aug 10 2017 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k += b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 07 2019 *)
    Table[j! Coefficient[Series[Exp[Exp[x] - 1], {x, 0, 20}], x, j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 01 2023 *)
    Table[(D[Exp[Exp[x]], {x, n}] /. x -> 0)/E, {n, 0, 20}] (* Joan Ludevid, Nov 05 2024 *)
  • Maxima
    makelist(belln(n),n,0,40); /* Emanuele Munarini, Jul 04 2011 */
    
  • PARI
    {a(n) = my(m); if( n<0, 0, m = contfracpnqn( matrix(2, n\2, i, k, if( i==1, -k*x^2, 1 - (k+1)*x))); polcoeff(1 / (1 - x + m[2,1] / m[1,1]) + x * O(x^n), n))}; /* Michael Somos */
    
  • PARI
    {a(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n)}; /* Michael Somos, Aug 22 2004 */
    
  • PARI
    a(n)=round(exp(-1)*suminf(k=0,1.0*k^n/k!)) \\ Gottfried Helms, Mar 30 2007 - WARNING! For illustration only: Gives silently a wrong result for n = 42 and an error for n > 42, with standard precision of 38 digits. - M. F. Hasler, Nov 30 2018
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) - 1), n))}; /* Michael Somos, Jun 28 2009 */
    
  • PARI
    Vec(serlaplace(exp(exp('x+O('x^66))-1))) \\ Joerg Arndt, May 26 2012
    
  • PARI
    A000110(n)=sum(k=0,n,stirling(n,k,2)) \\ M. F. Hasler, Nov 30 2018
    
  • Perl
    use bignum;sub a{my($n)=@;my@t=map{[(0)x($n+1)]}0..$n;$t[0][0]=1;for my$i(1..$n){$t[$i][0]=$t[$i-1][$i-1];for my$j(1..$i){$t[$i][$j]=$t[$i-1][$j-1]+$t[$i][$j-1]}}return map{$t[$][0]}0..$n-1}print join(", ",a(28)),"\n" # Paul Muljadi, May 08 2024
  • Python
    # The objective of this implementation is efficiency.
    # m -> [a(0), a(1), ..., a(m)] for m > 0.
    def A000110_list(m):
        A = [0 for i in range(m)]
        A[0] = 1
        R = [1, 1]
        for n in range(1, m):
            A[n] = A[0]
            for k in range(n, 0, -1):
                A[k-1] += A[k]
            R.append(A[0])
        return R
    A000110_list(40) # Peter Luschny, Jan 18 2011
    
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A000110, blist, b = [1,1], [1], 1
    for _ in range(20):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A000110.append(b) # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 19 2014
    
  • Python
    from sympy import bell
    print([bell(n) for n in range(27)]) # Michael S. Branicky, Dec 15 2021
    
  • Python
    from functools import cache
    @cache
    def a(n, k=0): return int(n < 1) or k*a(n-1, k) + a(n-1, k+1)
    print([a(n) for n in range(27)])  # Peter Luschny, Jun 14 2022
    
  • Sage
    from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    [bell_number(n) for n in (0..40)] # G. C. Greubel, Jun 13 2019
    

Formula

E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n-1} (1/(n-1)!)*A000166(j)*binomial(n-1, j)*(n-j)^(n-1). - André F. Labossière, Dec 01 2004
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/exp(1) = hypergeom([-1/x], [(x-1)/x], 1)/exp(1) = ((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/x)), where LaguerreL(mu, nu, z) = (gamma(mu+nu+1)/(gamma(mu+1)*gamma(nu+1)))* hypergeom([-mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. - Karol A. Penson, Mar 25 2002
a(n) = exp(-1)*Sum_{k >= 0} k^n/k! [Dobinski]. - Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(b-n-1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493). - Benoit Cloitre, Oct 23 2002, corrected by Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, North-Holland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz. - N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} (1-j*x)) (see Klazar for a proof). - Ralf Stephan, Apr 18 2004
a(n+1) = exp(-1)*Sum_{k>=0} (k+1)^(n)/k!. - Gerald McGarvey, Jun 03 2004
For n>0, a(n) = Aitken(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (A011971)]. - Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*binomial(k, i)*i^n + 0^n). - Paul Barry, Apr 18 2005
a(n) = A032347(n) + A040027(n+1). - Jon Perry, Apr 26 2005
a(n) = (2*n!/(Pi*e))*Im( Integral_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. - David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*x^2/(...)))))) (continued fraction due to Ph. Flajolet). - Paul D. Hanna, Jan 17 2006
From Karol A. Penson, Jan 14 2007: (Start)
Representation of Bell numbers B(n), n=1,2,..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2,..., i.e., having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(-1)*hypergeom([],[],1)=1
B(2)=exp(-1)*hypergeom([2],[1],1)=2
B(3)=exp(-1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(-1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). - Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n-1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta). - Jonathan Vos Post, Aug 27 2007
From Tom Copeland, Oct 10 2007: (Start)
a(n) = T(n,1) = Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * Lag(n,-1,j-n) = Sum_{j=0..n} [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0..n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,-1, j-n)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,-1, j-n)].
(End)
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then for Bell numbers B_n we have B_n=1/e*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) where a(0)=1. - Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(k-r). - David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k. - N. J. A. Sloane, Feb 07 2009)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] - R. J. Mathar, Jun 24 2024
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]:=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jul 08 2010
G.f. satisfies A(x) = (x/(1-x))*A(x/(1-x)) + 1. - Vladimir Kruchinin, Nov 28 2011
G.f.: 1 / (1 - x / (1 - 1*x / (1 - x / (1 - 2*x / (1 - x / (1 - 3*x / ... )))))). - Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 = A048993. - Wolfdieter Lang, Feb 03 2015
G.f.: (-1)^(1/x)*((-1/x)!/e + (!(-1-1/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments. - Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011 - Oct 2013 by Sergei N. Gladkovskii: (Start)
E.g.f.: exp(exp(x)-1) = 1 + x/(G(0)-x); G(k) = (k+1)*Bell(k) + x*Bell(k+1) - x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/G(0) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: -G(0) where G(k) = 1 - (x*k - 2)/(x*k - 1 - x*(x*k - 1)/(x + (x*k - 2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2 - (2*x*k - 1)/(x*k - 1 - x*(x*k - 1)/(x + (2*x*k - 1)/G(k+1) )); (continued fraction).
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 + x/( x*k - 1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k) = 1 - x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. - Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). - Vaclav Kotesovec, Nov 13 2015
From Natalia L. Skirrow, Apr 13 2025: (Start)
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
a(n) are the coefficients in the asymptotic expansion of -exp(-1)*(-1)^x*x*Gamma(-x,0,-1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function. - Vladimir Reshetnikov, Nov 12 2015
a(n) = 1 + floor(exp(-1) * Sum_{k=1..2*n} k^n/k!). - Vladimir Reshetnikov, Nov 13 2015
a(p^m) == m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference). - Seiichi Manyama, Jun 01 2016
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n} A182386(k)*Stirling2(n+1, k+1). - Mélika Tebni, Jul 02 2022
For n >= 1, a(n) = Sum_{i=0..n-1} a(i)*A074664(n-i). - Davide Rotondo, Apr 21 2024
a(n) is the n-th derivative of e^e^x divided by e at point x=0. - Joan Ludevid, Nov 05 2024

Extensions

Edited by M. F. Hasler, Nov 30 2018

A007489 a(n) = Sum_{k=1..n} k!.

Original entry on oeis.org

0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113, 4037913, 43954713, 522956313, 6749977113, 93928268313, 1401602636313, 22324392524313, 378011820620313, 6780385526348313, 128425485935180313, 2561327494111820313, 53652269665821260313, 1177652997443428940313
Offset: 0

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Author

Keywords

Comments

Equals row sums of triangle A143122 starting (1, 3, 9, 33, ...). - Gary W. Adamson, Jul 26 2008
a(n) for n>=4 is never a perfect square. - Alexander R. Povolotsky, Oct 16 2008
Number of cycles that can be written in the form (j,j+1,j+2,...), in all permutations of {1,2,...,n}. Example: a(3)=9 because in (1)(2)(3), (1)(23), (12)(3), (13)(2), (123), (132) we have 3+2+2+1+1+0=9 such cycles. - Emeric Deutsch, Jul 14 2009
Conjectured to be the length of the shortest word over {1,...,n} that contains each of the n! permutations as a factor (cf. A180632) [see Johnston]. - N. J. A. Sloane, May 25 2013
The above conjecture has been disproven for n>=6. See A180632 and the Houston 2014 reference. - Dmitry Kamenetsky, Mar 07 2016
a(n) is also the number of compositions of n if cardinal values do not matter but ordinal rankings do. Since cardinal values do not matter, a sequence of k summands summing to n can be represented as (s(1),...,s(k)), where the s's are positive integers and the numbers in parentheses are the initial ordinal rankings. The number of compositions of these summands are equal to k!, with k ranging from 1 to n. - Gregory L. Simay, Jul 31 2016
When the numbers denote finite permutations (as row numbers of A055089) these are the circular shifts to the left. Compare array A211370 for circular shifts to the left in a broader sense. Compare sequence A001563 for circular shifts to the right. - Tilman Piesk, Apr 29 2017
Since a(n) = (1!+2!+3!+...+n!) = 3(1+3!/3+4!/3+...+n!/3) is a multiple of 3 for n>2, the only prime in this sequence is a(2) = 3. - Eric W. Weisstein, Jul 15 2017
Generalization of 2nd comment: a(n) for n>=4 is never a perfect power (A007916) (Chentzov link). - Bernard Schott, Jan 26 2023

Examples

			a(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. - _Michael B. Porter_, Aug 03 2016
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Section B44, Springer 2010.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals A003422(n+1) - 1.
Column k=0 of A120695.

Programs

Formula

a(n) = Sum_{k=1..n} P(n, k)/C(n, k). - Ross La Haye, Sep 21 2004
a(n) = 3*A056199(n) for n>=2. - Philippe Deléham, Feb 10 2007
a(n) = !(n+1)-1=A003422(n+1)-1. - Artur Jasinski, Nov 08 2007 [corrected by Werner Schulte, Oct 20 2021]
Starting (1, 3, 9, 33, 153, ...), = row sums of triangle A137593 - Gary W. Adamson, Jan 28 2008
a(n) = a(n-1) + n! for n >= 1. - Jaroslav Krizek, Jun 16 2009
E.g.f. A(x) satisfies to the differential equation A'(x)=A(x)+x/(1-x)^2+1. - Vladimir Kruchinin, Jan 22 2011
a(0)=0, a(1)=1, a(n) = (n+1)*a(n-1)-n*a(n-2). - Sergei N. Gladkovskii, Jul 05 2012
G.f.: W(0)*x/(2-2*x) , where W(k) = 1 + 1/( 1 - x*(k+2)/( x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
G.f.: x /(1-x)/Q(0),m=+2, where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) ; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
E.g.f.: exp(x-1)*(Ei(1) - Ei(1-x)) - exp(x) + 1/(1 - x), where Ei(x) is the exponential integral. - Ilya Gutkovskiy, Nov 27 2016
a(n) = sqrt(a(n-1)*a(n+1)-a(n-2)*n*n!), n >= 2. - Gary Detlefs, Oct 26 2020
a(n) ~ n!. - Ridouane Oudra, Jun 11 2025

A005165 Alternating factorials: n! - (n-1)! + (n-2)! - ... 1!.

Original entry on oeis.org

0, 1, 1, 5, 19, 101, 619, 4421, 35899, 326981, 3301819, 36614981, 442386619, 5784634181, 81393657019, 1226280710981, 19696509177019, 335990918918981, 6066382786809019, 115578717622022981, 2317323290554617019, 48773618881154822981
Offset: 0

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Author

Keywords

Comments

Conjecture: for n > 2, smallest prime divisor of a(n) > n. - Gerald McGarvey, Jun 19 2004
Rebuttal: This is not true; see Zivkovic link (Math. Comp. 68 (1999), pp. 403-409) has demonstrated that 3612703 divides a(n) for all n >= 3612702. - Paul Jobling, Oct 18 2004
Conjecture: For n>1, a(n) is the number of lattice paths from (0,0) to (n+1,0) that do not cross above y=x or below the x-axis using up-steps +(1,a) and down-steps +(1,-b) where a and b are positive integers. For example, a(3) = 5: [(1,1)(1,1)(1,1)(1,-3)], [(1,1)(1,-1)(1,3)(1,-3)], [(1,1)(1,-1)(1,2)(1,-2)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,-1)(1,-1)]. - Nicholas Ham, Aug 23 2015
Ham's claim is true for n=2. We proceed with a proof for n>2 by induction. On the j-th step, from (j-1,y) to (j,y'), there are j options for y': 0, 1, ..., y-1, y+1, ..., j. Thus there are n! possible paths from (0,0) to x=n that stay between y=0 and y=x. (Then the final step is determined.) However, because +(1,0) is not an allowable step, we cannot land on (n,0) on the n-th step. Therefore, the number of acceptable lattice paths is n! - a(n-1). - Danny Rorabaugh, Nov 30 2015

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B10, pp. 152-153.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..30],n->Sum([1..n],i->(-1)^(n-i)*Factorial(i))); # Muniru A Asiru, Jun 01 2018
  • Haskell
    a005165 n = a005165_list !! n
    a005165_list = 0 : zipWith (-) (tail a000142_list) a005165_list
    -- Reinhard Zumkeller, Jul 21 2013
    
  • Maple
    A005165 := proc(n) local i; add((-1)^(n-i)*i!,i=1..n); end;
  • Mathematica
    nn=25;With[{fctrls=Range[nn]!},Table[Abs[Total[Times@@@Partition[ Riffle[ Take[ fctrls,n],{1,-1}],2]]],{n,nn}]] (* Harvey P. Dale, Dec 10 2011 *)
    a[0] = 0; a[n_] := n! - a[n - 1]; Array[a, 26, 0] (* Robert G. Wilson v, Aug 06 2012 *)
    RecurrenceTable[{a[n] == n! - a[n - 1], a[0] == 0}, a, {n, 0, 20}] (* Eric W. Weisstein, Jul 27 2017 *)
    AlternatingFactorial[Range[0, 20]] (* Eric W. Weisstein, Jul 27 2017 *)
    a[n_] = (-1)^n (Exp[1]((-1)^n Gamma[-1-n,1] Gamma[2+n] - ExpIntegralEi[-1]) - 1)
    Table[a[n] // FullSimplify, {n, 0, 20}] (* Gerry Martens, May 22 2018 *)
  • PARI
    a(n)=if(n<0,0,sum(k=0,n-1,(-1)^k*(n-k)!))
    
  • PARI
    first(m)=vector(m,j,sum(i=0,j-1,((-1)^i)*(j-i)!)) \\ Anders Hellström, Aug 23 2015
    
  • PARI
    a(n)=round((-1)^n*(exp(1)*(gamma(n+2)*incgam(-1-n,1)*(-1)^n +eint1(1))-1)) \\ Gerry Martens, May 22 2018
    
  • Python
    a = 0
    f = 1
    for n in range(1, 33):
        print(a, end=",")
        f *= n
        a = f - a
    # Alex Ratushnyak, Aug 05 2012
    

Formula

a(0) = 0, a(n) = n! - a(n-1) for n > 0; also a(n) = n*a(n-2) + (n-1)*a(n-1) for n > 1. Sum_{n>=1} Pi^n/a(n) ~ 30.00005. - Gerald McGarvey, Jun 19 2004
E.g.f.: 1/(1-x) + exp(-x)*(e*(Ei(1,1)-Ei(1,1-x)) - 1). - Robert Israel, Dec 01 2015
a(n) = (-1)^n*(exp(1)*(gamma(n+2)*gamma(-1-n,1)*(-1)^n +Ei(1))-1). - Gerry Martens, May 22 2018
Sum_{n>=1} 1/a(n) = A343187. - Amiram Eldar, Jun 01 2023
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